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I When to use which dimensionless number

  1. Dec 15, 2017 #41
    I get $$\frac{\sqrt{g_{zz}}}{\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}}$$
     
  2. Dec 16, 2017 #42

    joshmccraney

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    Ok, so we have:
    $$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z \implies\\
    |\hat{a}^\theta| = \sqrt{\left(\frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\right)^2+\left(\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\right)^2}\\
    =\sqrt{\frac{g_{zz}^2}{(-g_{\theta z}^2+g_{zz}g_{\theta\theta})^2}+\frac{g_{\theta z}^2}{(g_{\theta z}^2-g_{zz}g_{\theta\theta})^2}}\\
    =\sqrt{\frac{g_{zz}^2+g_{\theta z}^2} {(-g_{\theta z}^2+g_{zz}g_{\theta\theta})^2}}$$
    What do you think?
     
  3. Dec 16, 2017 #43
    I think that ##\hat{a}_{\theta}## and ##\hat{a}_{z}## are not orthogonal, nor are they unit vectors.
     
  4. Dec 16, 2017 #44

    joshmccraney

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    Riiiiiiight, as you said bevore $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}$$
    My bad. So I would take the norm by transforming back to these definitions. Somehow I have a feeling you have a more general approach? Could you share it and I'll do the footwork? I'd like to see how to take a vector norm with non-orthonormal basis vectors.
     
  5. Dec 16, 2017 #45
    Just use Eqns. 5 in post #35.
     
  6. Jan 3, 2018 #46

    joshmccraney

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    I tried thinking about how equations 5 help, but I can't see how. At any rate, we have
    $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$
    $$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}$$
    $$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z \\
    =\frac{-g_{zz}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \left( \frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta} \right)+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\left( \frac{\partial R}{\partial z}\hat{r}+\hat{z} \right)\\
    = \frac{g_{\theta z} \partial_zR -g_{zz} \partial_\theta R}{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \hat{r} +
    \frac{-g_{zz} R}{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \hat{\theta}
    + \frac{g_{\theta z} }{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \hat{z}$$
    where from here I could compute the norm as a regular problem. Would you mind highlighting the procedure you referenced, involving equations 5? Again, I'm sorry for the amount of time it's taken for me to reply!
     
  7. Jan 3, 2018 #47
    $$\hat{a}^{\theta}=\frac{g_{zz}\hat{a}_{\theta}-g_{\theta z}\hat{a}_z}{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$$
    $$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{(g_{zz}\hat{a}_{\theta}-g_{\theta z}\hat{a}_z)\centerdot (g_{zz}\hat{a}_{\theta}-g_{\theta z}\hat{a}_z)}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
    $$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}^2(\hat{a}_{\theta}\centerdot \hat{a}_{\theta})-2g_{\theta z}g_{zz}(\hat{a}_z\centerdot \hat{a}_{\theta})+g_{\theta z}^2(\hat{a}_z\centerdot \hat{a}_z)}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
    $$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}^2g_{\theta \theta}-2g_{\theta z}^2g_{zz}+g_{\theta z}^2g_{zz}}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
    $$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}(g_{zz}g_{\theta \theta}-g_{\theta z}^2)}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
    $$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}}{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$$
     
  8. Jan 3, 2018 #48

    joshmccraney

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    Riiiiiight, the magnitude of any vector ##\vec v =\sqrt{ \vec v \cdot \vec v}## since ##\vec v \cdot \vec v = |v| |v| \cos \theta = |v|^2##. Shoot, this makes perfect sense now! Okay, so I agree the magnitude is $$\frac{\sqrt{g_{zz}}}{\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}}.$$ From the above we know
    $$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z\implies\\
    \hat u^\theta = \sqrt{g_{zz}}\hat{a}_\theta - \frac{g_{\theta z}}{\sqrt{g_{zz}}}\hat{a}_z\\
    =\frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}}.$$
    How does that look?
     
  9. Jan 3, 2018 #49
    I think you already had the equation for ##\hat{a}^{\theta}##. If you're trying to represent the unit vector in the same direction as ##\hat{a}^{\theta}## (which this is not), then at you're missing a factor of ##\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}## in the denominator.
     
  10. Jan 3, 2018 #50

    joshmccraney

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    You previously posted
    We found ####\hat{a}^{\theta}## and its magnitude, right? What I showed was the division. Am I missing something (perhaps messed up the bookkeeping)?
     
  11. Jan 3, 2018 #51
    We are looking for the unit vector in the same direction as a super theta. You can’t just divide a vector by its magnitude and use the same symbol for it. Plus, if you are dividing a super theta by its own magnitude to get the unit vector, you need to do the algebra correctly and you need to use a new symbol for it.
     
  12. Jan 3, 2018 #52

    joshmccraney

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    In post 37 you say:
    and in post 47 you show how to compute the magnitude of ##\hat{a}^{\theta}##. What I tried doing in post 48 was dividing ##\hat{a}^{\theta}## by it's magnitude, making it a unit vector.

    Also, did I use the same symbol? Thought I used your suggestion ##\hat{u}^{\theta}##.
     
  13. Jan 3, 2018 #53

    joshmccraney

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    Yea, I dropped a term in the algebra. We should have
    $$
    \hat u^\theta
    =\frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}.$$
     
  14. Jan 3, 2018 #54
    Right. Now, by extension, please write down what ##\hat{u}^z## would be.
     
  15. Jan 3, 2018 #55

    joshmccraney

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    Without showing any of the legwork since it's the same as your approach, we have $$\hat{u}^z = \frac{g_{\theta\theta} \hat a_z-g_{\theta z}\hat a_\theta}{\sqrt{g_{\theta \theta}}\sqrt{g_{zz}g_{\theta\theta}-g_{\theta z}^2}}.$$
    Do you agree?
     
  16. Jan 4, 2018 #56
    Yes. I'll be back later to continue.
     
  17. Jan 4, 2018 #57
    To refresh your memory, we are going to be performing a force balance over a "window" in the free surface formed by the intersection of planes of constant ##\theta## and z with our free surface. These intersections form grid lines on our free surface. The window will be situated between z and z + dz, and between ##\theta## and ##\theta+d\theta##. In terms of ##g_{zz}##, ##g_{\theta z}##, ##g_{\theta \theta}##, dz, and ##d\theta##, what are the lengths of the four sides of our window?
     
  18. Jan 12, 2018 #58

    joshmccraney

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    I'm not quite sure what to write. I'm trying to compare what we have to a simpler problem. Geometrically, if we were in cylindrical coordinates the side lengths would be ##r\Delta\theta,\Delta r,\Delta r,(r+\Delta r)\Delta \theta##. This makes me think the lengths should be something like ##a \Delta z, b \Delta z, c \Delta \theta, d \Delta \theta##. The ##g##-subs (as oppose to -supers) were dot products of the in plane vectors ##a##-subs (not necessarily orthonormal). I'm stuck; any hints?

    To make sure I understand what we've done, we started by finding two vectors that lie tangent in the plane and call them ##a##-sub. Then we carefully aligned them with our selected coordinate system and called them ##a##-super. Then we normalized them and called them ##u##. Right?
     
  19. Jan 13, 2018 #59
    We found two vectors in the tangent plane that align with the curved grid lines of constant theta and constant z within the surface. These vectors, when multiplied by the corresponding differentials in the coordinates represent actual differential position vectors within the surface in terms of both their spatial length and their direction. Then we found the a-super vectors normal to the grid lines, with the characteristic that, when dotted with the subs, the dot product is equal to 1 or zero. We then used this to find the equations for unit vectors in the plane of the surface that are normal to the grid lines.

    Regarding our little window in the surface, the sides of the window have differential position vectors along their edges (from corner to corner) of ##\mathbf{a}_{\theta}d\theta## and ##\mathbf{a}_{z}dz##. So, what are the lengths of these edges?
     
  20. Jan 14, 2018 #60

    joshmccraney

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    Would the sides be ##\sqrt{g_{\theta \theta}}|_zd\theta##, ##\sqrt{g_{\theta \theta}}|_{z+dz}d\theta##, ##\sqrt{g_{zz}}|_\theta dz ##, ##\sqrt{g_{zz}}|_{\theta+d\theta}dz##?
     
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